Sequentially compact

[Imum Coeli]

Hi I'm stuck on a proof and don't know what to do.

Q: A set \(\displaystyle S \subseteq \mathbb{R} \) is said to be sequentially compact if every sequence in \(\displaystyle S \) has a subsequence that converges to a point in \(\displaystyle S\). Prove that \(\displaystyle S\) is sequentially compact iff \(\displaystyle S\) is closed and bounded.

A:
Will first prove the reverse implication.Suppose that \(\displaystyle S \) is closed and bounded.
Since \(\displaystyle S \) is bounded then every sequence in \(\displaystyle S \) is bounded, so by the Bolzano-Weierstrass theorem every sequence in \(\displaystyle S\) has a convergent subsequence.
Also, since \(\displaystyle S\) is closed, then the limit of any convergent sequence is in \(\displaystyle S\)
Therefore \(\displaystyle S\) is sequentially compact


Now will (try to) prove the forward implication.Suppose that \(\displaystyle S \) is sequentially compact.
Then every sequence in \(\displaystyle S \) has a subsequence that converges to a point in \(\displaystyle S\), so... And I have nothing.


All I can think of is that the convergent subsequences are closed and bounded but I don't see how that helps. Any advice.

 

[Ishuda]

I haven't really looked at it yet but, off the top of my head, it looks like a Reductio ad absurdum type argument might work. That is assume that [FONT=MathJax_Math]S[/FONT] is sequentially compact and it is not the case that [FONT=MathJax_Math]S[/FONT] is closed and bounded. Now show this leads to an absurd/ridiculous/incorrect conclusion

 

[Imum Coeli]

Okay after an epiphany I decided to try for a contradiction.

Let \(\displaystyle S \) be sequentially compact.
Suppose that \(\displaystyle S \) is neither closed nor bounded.

Since \(\displaystyle S \) is open it has a sequence, \(\displaystyle (x_n)\), that converges to some \(\displaystyle s_o\; \notin\; S\)
\(\displaystyle \implies \) every subsequence of \(\displaystyle (x_n)\) converges to \(\displaystyle s_o\; \notin \; S\).
\(\displaystyle \implies \; \exists \; \) a subsequnce in \(\displaystyle S\) that does not converge to a point in \(\displaystyle S. \; \Rightarrow|\Leftarrow \)
Therefore \(\displaystyle S\) is closed.

Also since \(\displaystyle S\) is unbounded \(\displaystyle \exists \) a sequence \(\displaystyle (x_n) \subseteq S : |x_n|\geq k\; \forall\; k\in \mathbb{N}\)
so \(\displaystyle (x_n) \) has no convergent subsequence. \(\displaystyle \Rightarrow|\Leftarrow \)
Therefore \(\displaystyle S\) is bounded.

 

[Imum Coeli]

I haven't really looked at it yet but, off the top of my head, it looks like a Reductio ad absurdum type argument might work. That is assume that [FONT=MathJax_Math]S[/FONT] is sequentially compact and it is not the case that [FONT=MathJax_Math]S[/FONT] is closed and bounded. Now show this leads to an absurd/ridiculous/incorrect conclusion

I just thought the same thing. Thanks.

 

[Ishuda]

Almost. It is not necessary nor true for all sequences, for example let S=(0,∞) and
x = {.5, .1,1, .5, .01,2,.5,.001,3,.5,.001,4, ..., .5, 10^-n, n, ...}.
Then x has a sub-sequence which converges to a point in S, a sub-sequence which converges to a point not in S, and an unbounded sub-sequence in S

You just need the one counter example to prove that is not true for every sequence.

 

[Imum Coeli]

Is this the problem?

"Since \(\displaystyle S \) is open it has a sequence, \(\displaystyle (x_n)\), that converges to some \(\displaystyle s_o\; \notin\; S\)
\(\displaystyle \implies \) every subsequence of \(\displaystyle (x_n)\) converges to \(\displaystyle s_o\; \notin \; S\)."

Isn't it true that if a sequence \(\displaystyle (x_n)\) converges to some \(\displaystyle L \in \mathbb{R} \) then every subsequence of \(\displaystyle (x_n)\) also converges to \(\displaystyle L\)?

 

[Ishuda]

Is this the problem?

"Since \(\displaystyle S \) is open it has a sequence, \(\displaystyle (x_n)\), that converges to some \(\displaystyle s_o\; \notin\; S\)
\(\displaystyle \implies \) every subsequence of \(\displaystyle (x_n)\) converges to \(\displaystyle s_o\; \notin \; S\)."

Isn't it true that if a sequence \(\displaystyle (x_n)\) converges to some \(\displaystyle L \in \mathbb{R} \) then every subsequence of \(\displaystyle (x_n)\) also converges to \(\displaystyle L\)?

Yes, that is correct. I'm sorry I got confused in what I was reading and thought you had said if x had a sub-sequence which converged to a point s not is S then every sub-sequence of x converged to the point s. A completely different (and incorrect) statement.

 

[Imum Coeli]

No worries. Thanks heaps for your help.